[Math] Do the probabilities in the expected value formula need to sum to 100%

gamblingprobabilitystatistics

The formula for expected value in my math textbook is as follows:
∑ x*P(X=x)

I got into a discussion with my professor about whether or not the probabilities used in the formula need to sum to 100%.

The problem that caused the discussion is:
Invent a gambling game where some money is paid for rolling 2 dice whose sum is 7, otherwise looses his money. How much should you pay for a 1 dollar bet, if you want the game to be fair?

So my logic was as follows:

  • You are guaranteed to loose a dollar. So the first event is -1*100%
  • Next getting a sum of seven has 1/6 chance. So the second event is x(1/6)
  • My math would be: -1 + x(1/6) = 0 => x=6

My professor did the following:

  • You either win or loose.

  • Odds of loosing a dollar 5/6. So the first event is -1*5/6

  • Odds of winning 1/6. So the second even is x*1/6

  • His math would be: -1(5/6) + x(1/6) = 0 => x=5

  • Then he adds a dollar at the end to pay back the initial bet.

We both got 6 for the amount of money returned.

My professor told me my math did not work since the probability of my event did not sum to 100%. However, the formula given never requires that the sum to 100%. I also want to mention that the odds that you either win or loose is 100%, since loosing a dollar happens whether or not you win.

Best Answer

You computed the expected value by a different summation; if you're actually computing the expected value of the random variable $X$ by looking at $x P(X = x)$, and summing, then the probabilities do have to add to 1, because that's the definition of expected value.

In your case, what's the random variable $X$? It's not "how much you win", is it? In fact, you've got TWO events.

$X_1$ = -1 for all dice rolls $X_2$ = +1/6 for rolls that total 7, 0 otherwise.

You've observed that $X = X_1 + X_2$, and then have computed the expected values of $X_1$ and $X_2$ separately.

Your computation for $X_1$ has a sum of $-1 \cdot 1$; the probabilities (there's only 1) sum to 1 in this case.

Your computataion for $X_2$ is similar, but looks like $x \cdot \frac{1}{6} + 0 \cdot \frac{5}{6}$, and again, the probabilities sum to 1.

Critical to this is the theorem that $$ E[X_1 + X_2] = E[X_1] + E[X_2] $$ which neither you nor your prof seems to have mentioned.

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